`=>` Let `L_1` and `L_2` be two lines passing through the origin and with direction ratios `a_1, b_1, c_1` and `a_2, b_2, c_2`, respectively.
`=>` Let `P` be a point on `L_1` and Q be a point on `L_2`. Consider the directed lines `OP` and `OQ` as given in Fig .
`=>` Let `θ` be the acute angle between `OP` and `OQ.` Now recall that the directed line segments `OP` and `OQ` are vectors with components `a_1, b_1, c_1` and `a_2, b_2, c_2`, respectively. Therefore, the angle `θ` between them is given by
`color{red}{cos theta= | (a_1 a_2 + b_1 b_2 + c_1 c_2 )/(sqrt (a_(1)^2 + b_(1)^2 + c_(1)^2) sqrt ( a_(2)^2 + b_(2)^2 + c_(2)^2) ) |}`
`=>` The angle between the lines in terms of sin θ is given by
`sin theta = sqrt(1- cos^2 theta)`
`= sqrt (1- ( (a_1 a_2 + b_1 b_2 + c_1 c_2 )^2)/( ( a_(1)^2 + b_(1)^2 + c_(1)^2) ( a_(2)^2 + b_(2)^2 + c_(2)^2) ) )`
`= ( sqrt ( ( a_(1)^2 + b_(1)^2 + c_(1)^2) ( a_(2)^2 + b_(2)^2 + c_(2)^2) - (a_(1) a_2 + b_1 b_2 + c_1 c_2)^2 ) )/(sqrt ( (a_(1)^2 + b_(1)^2 + c_(1)^2) ) sqrt (a_(2)^2 + b_(2)^2 + c_(2)^2) )`
`color{red}{sintheta= ( sqrt ( (a_1 b_2 - a_2 b_1)^2 + ( b_1 c_2 - b_2 c_1)^2 + ( c_1 a_2 - c_2 a_1)^2 ) )/( sqrt (a_(1)^2 + b_(1)^2 + c_(1)^2) sqrt ( a_(2)^2 + b_(2)^2 + c_(2)^2) )}` ................(2)
`color{red} "Key Point"` - In case the lines `L_1` and `L_2` do not pass through the origin, we may take lines `L'_(1)` and `L′_(2)` which are parallel to `L_1` and `L_2` respectively and pass through the origin.
`\color{green} ✍️` If instead of direction ratios for the lines `L_1` and `L_2`, direction cosines, namely, `l_1, m_1, n_1` for `L_1` and `l_2, m_2, n_2` for `L_2` are given, then (1) and (2) takes the following form:
`color{red}{cos theta = | l_1 l_2 + m_1 m_2 + n_1 n_2 |}`
(as `l_(1)^2 + m_(1)^2 + n_(1)^2 =1 = l_(2)^2 + m_(2)^2 + n_(2)^2)` .....(3)
and `sin theta = sqrt ( (l_1 m_2 - l_2 m_1 )^2 - (m_1 n_2 - m_2 n_1)^2 + (n_1 l_2 - n_2 l_1)^2 )` ....(4)
`=>` Two lines with direction ratios `a_1, b_1, c_1` and `a_2, b_2, c_2` are
(i) perpendicular i.e. if `θ = 90° `by (1)
`color{blue}{a_1 a_2 + b_1 b_2 + c_1 c_2 = 0}`
(ii) parallel i.e. if `θ = 0` by (2)
`color{blue}{a_1/a_2 =b_1/b_2 =c_1/c_2}`
`\color{green} ✍️` Now, we find the angle between two lines when their equations are given. If θ is acute the angle between the lines
`vec r = bar (a_1) + lambda bar (b_1)` and `vec r = bar (a_2) + mu bar ( b_2)`
then `cos theta= | ( bar (b_1) * bar (b_2) )/( | bar (b_1) | | bar (b_2) | ) |`
`\color{green} ✍️` So In Cartesian form, if θ is the angle between the lines
` (x-x_1)/(a_1) = (y-y_1)/(b_1) = (z-z_1)/(c_1)` ........(1)
and `(x-x_2)/(a_2) = (y-y_2)/(b_2) = (z-z_2)/(c_2)` ........(2)
where, `a_1, b_1, c_1` and `a_2, b_2, c_2` are the direction ratios of the lines (1) and (2), respectively, then
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \color{red} {cos theta = | (a_1 a_2 + b_1 b_2 + c_1 c_2)/( sqrt (a_(1)^2 + b_(1)^2 + c_(1)^2) sqrt (a_(2)^2 + b_(2)^2 + c_(2)^2) ) |}`
`=>` Let `L_1` and `L_2` be two lines passing through the origin and with direction ratios `a_1, b_1, c_1` and `a_2, b_2, c_2`, respectively.
`=>` Let `P` be a point on `L_1` and Q be a point on `L_2`. Consider the directed lines `OP` and `OQ` as given in Fig .
`=>` Let `θ` be the acute angle between `OP` and `OQ.` Now recall that the directed line segments `OP` and `OQ` are vectors with components `a_1, b_1, c_1` and `a_2, b_2, c_2`, respectively. Therefore, the angle `θ` between them is given by
`color{red}{cos theta= | (a_1 a_2 + b_1 b_2 + c_1 c_2 )/(sqrt (a_(1)^2 + b_(1)^2 + c_(1)^2) sqrt ( a_(2)^2 + b_(2)^2 + c_(2)^2) ) |}`
`=>` The angle between the lines in terms of sin θ is given by
`sin theta = sqrt(1- cos^2 theta)`
`= sqrt (1- ( (a_1 a_2 + b_1 b_2 + c_1 c_2 )^2)/( ( a_(1)^2 + b_(1)^2 + c_(1)^2) ( a_(2)^2 + b_(2)^2 + c_(2)^2) ) )`
`= ( sqrt ( ( a_(1)^2 + b_(1)^2 + c_(1)^2) ( a_(2)^2 + b_(2)^2 + c_(2)^2) - (a_(1) a_2 + b_1 b_2 + c_1 c_2)^2 ) )/(sqrt ( (a_(1)^2 + b_(1)^2 + c_(1)^2) ) sqrt (a_(2)^2 + b_(2)^2 + c_(2)^2) )`
`color{red}{sintheta= ( sqrt ( (a_1 b_2 - a_2 b_1)^2 + ( b_1 c_2 - b_2 c_1)^2 + ( c_1 a_2 - c_2 a_1)^2 ) )/( sqrt (a_(1)^2 + b_(1)^2 + c_(1)^2) sqrt ( a_(2)^2 + b_(2)^2 + c_(2)^2) )}` ................(2)
`color{red} "Key Point"` - In case the lines `L_1` and `L_2` do not pass through the origin, we may take lines `L'_(1)` and `L′_(2)` which are parallel to `L_1` and `L_2` respectively and pass through the origin.
`\color{green} ✍️` If instead of direction ratios for the lines `L_1` and `L_2`, direction cosines, namely, `l_1, m_1, n_1` for `L_1` and `l_2, m_2, n_2` for `L_2` are given, then (1) and (2) takes the following form:
`color{red}{cos theta = | l_1 l_2 + m_1 m_2 + n_1 n_2 |}`
(as `l_(1)^2 + m_(1)^2 + n_(1)^2 =1 = l_(2)^2 + m_(2)^2 + n_(2)^2)` .....(3)
and `sin theta = sqrt ( (l_1 m_2 - l_2 m_1 )^2 - (m_1 n_2 - m_2 n_1)^2 + (n_1 l_2 - n_2 l_1)^2 )` ....(4)
`=>` Two lines with direction ratios `a_1, b_1, c_1` and `a_2, b_2, c_2` are
(i) perpendicular i.e. if `θ = 90° `by (1)
`color{blue}{a_1 a_2 + b_1 b_2 + c_1 c_2 = 0}`
(ii) parallel i.e. if `θ = 0` by (2)
`color{blue}{a_1/a_2 =b_1/b_2 =c_1/c_2}`
`\color{green} ✍️` Now, we find the angle between two lines when their equations are given. If θ is acute the angle between the lines
`vec r = bar (a_1) + lambda bar (b_1)` and `vec r = bar (a_2) + mu bar ( b_2)`
then `cos theta= | ( bar (b_1) * bar (b_2) )/( | bar (b_1) | | bar (b_2) | ) |`
`\color{green} ✍️` So In Cartesian form, if θ is the angle between the lines
` (x-x_1)/(a_1) = (y-y_1)/(b_1) = (z-z_1)/(c_1)` ........(1)
and `(x-x_2)/(a_2) = (y-y_2)/(b_2) = (z-z_2)/(c_2)` ........(2)
where, `a_1, b_1, c_1` and `a_2, b_2, c_2` are the direction ratios of the lines (1) and (2), respectively, then
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \color{red} {cos theta = | (a_1 a_2 + b_1 b_2 + c_1 c_2)/( sqrt (a_(1)^2 + b_(1)^2 + c_(1)^2) sqrt (a_(2)^2 + b_(2)^2 + c_(2)^2) ) |}`